Err...
Allowing that Monty is constrained to NOT opening the Door with the prize.
If the Prize were behind “A” - he could have opened either “B” or “C”. Thus, he was not “forced” to open “B” - “B” is simply a valid choice of available options.
The author concedes this when he states:
“Now suppose you had first picked C. Which doors could Monty have opened? Right: A or B. And in this case you should stay.”
The problem with the hypothesis is that the author, in his conclusions includes assumptions as evidence!
He asks - assume the prize is behind “C”, then “A” = 1/3, “C” = 2/3.
The problem is - at the time you need to make the decision, you have no evidence that the prize is behind “C” - you only know that it “could” be behind it, but that was a given from the beginning.
Right. You the contestant has no idea if Monty’s choice is constrained or not. Just that there is a 2/3 chance that it is. Based on this lack of evidence, there is never any reason to switch.
The author’s assumption that the prize is behind C invalidates his argument. Given that the prize is behind a closed door, there is no change in the equality of distribution of probability that the prize is behind a specific closed door, no matter how many other doors are opened.
If there are 5 doors, the chance that the prize is behind the first door is 1 in 5.
Open the fifth door, and the probability that the prize hides behind door number one is 1 in 4.
Open the fourth, 1 in 3.
The equal distribution of probability does not change, although the actual value of the probability does rise as each door is opened.
To take the argument presented here, one must assume that the initial pick was incorrect, with no evidence to support that assumption.